Giant Swing, Uniform Circular Motion

  1. Feb 12, 2008 #1
    [SOLVED] Giant Swing, Uniform Circular Motion

    1. The problem statement, all variables and given/known data

    The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 long, the upper end of the cable being fastened to the arm at a point 3.00 from the central shaft.

    [​IMG]

    A) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of with the vertical.

    B) Does the angle depend on the weight of the passenger for a given rate of revolution?


    2. Relevant equations

    R = Lsin(\theta)

    v = [tex]\sqrt{gtan(\theta)R}[/tex]

    T = 2[tex]\Pi[/tex]R/v


    3. The attempt at a solution

    tried using L = 3+5m*sin([tex]\theta[/tex])) to get 4m

    R then equals = 2m

    v then equals = [tex]\sqrt{9.8*2*tan\theta}[/tex] = 11.3m/s

    T then equals 2[tex]\Pi[/tex](2m)/11.3m/s = 1.1s wrong

    I tried a few other combinations where I used 3+(5sintheta) as the L and got T=4.4s which was wrong as well.

    I'm not sure what I'm doing wrong, I think I have the correct equations and I know I have the right angle and distances, I guess I'm just not sure how to derive the right length and and radius.

    I'm also thinking that for part B the weight will determine the angle that the seat swings, but I don't want to risk losing the only chance I have on that part of the problem (masteringphysics).

    any and all help is greatly appreciated.
     
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2
    You might want to go back and look at this bit again:
    sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.
     
  4. Feb 12, 2008 #3

    PhanthomJay

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    It appears that you are confusing L with R. If L is the length of the cable, then what is R, which is the radius measured from the passenger to the central shaft? Also, if you draw a free body diagram and use Newton's laws, it will help to see whether or not the the mass comes into play. You shouldn't blindly be using a formula..
     
  5. Feb 12, 2008 #4
    I wrote it before, R = Lsin(30)

    I was pretty sure it was the length of the cable alone (which would've been 5) I just wasn't sure where the distance of the angle of the seat from the vertical shaft (3m) played in.

    if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct?

    on the angle's dependance on the weight, I've drawn a free body diagram and it makes the sum of the forces be

    (sigma)Fx = -marad + mgcos30 = 0

    (sigma)Fy = T - mgsin30 = 0

    so I would imagine the angle is dependant on the weight of the seat.
     
  6. Feb 12, 2008 #5
    As I already said:
     
  7. Feb 12, 2008 #6
    thanks, I got the answer to the first part using that radius, I wasn't picturing in my head that the swing literally makes a circle when viewed overhead and the distance of 3m + (5sin30) was the radius of that circle.

    now fro for the 2nd part, as I said I did a free body diagram and found the forces acting in each direction, but I'm actually still not certain, I believe the mass would cancel out from the 2 equations if I set them equal to each other, so the answer would be that it doesn't depend on the mass?
     
  8. Feb 12, 2008 #7

    PhanthomJay

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    Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.
     
  9. Feb 12, 2008 #8
    so then in that case Tcos30 = mg

    so the angle is proportional to the mass?
     
  10. Feb 12, 2008 #9
    scracth that the angle is not proportionate to the mass, thanks for the help.
     
  11. Feb 28, 2008 #10
    I have the same problem as well, and I don't quite understand how velocity was derived.
     
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